# A problem with the path of least resistance

I have a question about short circuits that I haven't been able to solve.

From this we know: The total voltage V is constant, and the total current I is inversely proportional to the total resistance R.

Due to this being a parallel circuit, the voltage of each branch is the same and is equal to the total voltage: $$V_{AB}=V_{CD}=V(\text {constant})$$

Or we can write this as: $$I_{AB}R_{AB}=I_{CD}R_{CD}=V$$

Since current is inversely proportional to the total resistance, when RCD decreases, ICD will increase. At the same time, the total current I will also increase. $$I=I_{AB}+I_{CD}=\frac{V}{R}$$

But IAB will not change, otherwise the total voltage will change too, but the total voltage is constant.

Now comes my question: If this is so, let's suppose RCD approaches 0, which means there will be a short circuit. It is said that all the current will travel through the path of least resistance, which means no current passes through AB and thus the bulb won't light up.

However, in theory, there is still a current IAB flowing through the bulb notwithstanding the decrease in RCD which only affects ICD right? Shouldn't the bulb still light up then?

How do we square these two facts?

PS: I understand that if there is internal resistance, what I said no longer holds. But what if we imagine there was no internal resistance, does it mean that the bulb will light up?

Your reasoning is sound, but taking $$R_{CD}$$ to zero in these circumstances is unenlightening. As $$R_{CD}$$ approaches zero the current through it will become larger and larger and, as you say, the current through $$R_{AB}$$ will stay the same. I think that's all we can usefully say about this 'ideal' case.

You could say that when $$R_{CD}$$ became zero the current through it would be infinite, and so would be the battery current (or, if you had a sense of humour, that the battery current would be infinity + 100 mA). But neither of these forms of words would add anything to our understanding.

Clearly in practice, the battery will have an internal resistance and won't maintain a terminal pd of 10 V. Wires will have resistance and may melt.

• The battery may even explode. Sep 15, 2021 at 9:56

"Now comes my question: If this is so, let's suppose RCD approaches 0, which means there will be a short circuit. It is said that all the current will travel through the path of least resistance, which means no current passes through AB, and thus the bulb won't light up." Note that there will still be a very small amount of current flowing through the bulb, based on the ratio between the resistance AB and CD. Theoretical explanations of short circuits assume that there is 0 resistance through CD, which would give rise to an infinite current through CD, and therefore, 0 current through AB and the lamp. However, in real-life situations, 0 resistance cannot be achieved, as there is still the resistance of the wires to be accounted for. Therefore, there will still be a certain amount of current through AB, which, according to your question, you assume, will light the bulb. However, the current through the bulb is infinitesimal, and the bulb will not light not due to the fact that there is 0 current flowing through the bulb, but due to the fact that the current is not enough for the bulb to light up, which is dependent on its power rating. I am no expert in this subject but these are my thoughts. Maybe this could help: https://www.quora.com/How-much-current-flows-during-a-short-circuit-infinite-or-zero

It is said that all the current will travel through the path of least resistance, which means no current passes through AB and thus the bulb won't light up.

That is a false statement. Current splits between parallel paths in proportion to the resistance of those paths.

If you apply a zero resistance short from C to D then no current will flow in the A-B connected branch because you will have shorted it out with a perfectly zero resistance short.